Divide the polynomials. The form of your answer should either be $p(x)$ or $p(x)+\dfrac{k}{x-6}$ where $p(x)$ is a polynomial and $k$ is an integer. $\dfrac{2x^3-13x^2+9x-16}{x-6}=$
Solution: Usually, there are many different ways to divide polynomials. Here, we will use the method of polynomial long division. $\begin{array}{r} 2x^2-\phantom{1}x+\phantom{1}3 \\ x-6|\overline{2x^3-13x^2+9x-16} \\ \mathllap{-(}\underline{2x^3-12x^2\phantom{+9x-16}\rlap )} \\ -x^2+9x-16 \\ \mathllap{-(}\underline{-x^2+6x\phantom{-16}\rlap )} \\ 3x-16 \\ \mathllap{-(}\underline{3x-18\rlap )} \\ 2 \end{array}$ We found that the quotient is $2x^2-x+3$ and the remainder is $2$ : $\dfrac{2x^3-13x^2+9x-16}{x-6}=2x^2-x+3+\dfrac{2}{x-6}$